Question

A 46 g piece of metal at 250C is placed in a calorimeter with 500. mL of water at 20C. If the final temperature was 24C, the specific heat of the metal is what?

Answer 1

Step-by-step explanation:
This is a thermo-equillibrium situation. We can use the equation
Loss of Heat of the Metal = Gain of Heat by the Water
−
Q
m
=
+
Q
w
Q
=
m
Δ
T
C
p
Q
=
Heat
m
=
mass
Δ
T
=
(
T
f
−
T
i
)

T
f
=
Final Temp
T
i
=
Initial Temp
C
P
=
Specific Heat
Metal
Water
m
=
17.5
g

T
f
=
30.0
o
C

T
i
=
125.0
o
C

C
P
=
x
Water
m
=
15.0
g

T
f
=
30.0
o
C

T
i
=
25.0
o
C

C
P
=
4.184
J
g
o
C
#
−
Q
m
=
+
Q
w

−
[
m
(
T
f
−
T
−
i
)
C
p
]
=
m
(
T
f
−
T
−
i
)
C
p
−
[
17.5
g
(
30
o
C
−
125
o
C
)
x
]
=
15
g
(
30
o
C
−
25
o
C
)
4.184
J
g
o
C
−
[
17.5
g
(
−
95
o
C
)
x
]
=
15
g
(
5
o
C
)
)
4.184
J
g
o
C
(
1662.5
g
o
C
)
x
=
313.8
J
1662.5
g
o
C
x
1662.5
g
o
C
=
313.8
J
1662.5
g
o
C
C
p
=
0.189
J
g
o
C